I had to solve

$$\sin(5x) - \sin (3x) = \sqrt 2 \; \cos(4x)$$

After working with the equation, I got
$$2\sin(x)\cos(4x) = \sqrt 2 \; \cos(4x)$$ with difference of sines formula.

I saw that, I need to check if $\cos(4x)=0$ is a solution, so I got $x= \frac{\pi}{8}$ and it worked, so my first two solutions are

$$x= \frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= -\frac{\pi}{8} + 2\pi n \space, n \in \Bbb Z $$

After that, I divided by $\cos(4x)$, getting $\sin(x) = \frac{\sqrt 2}{2}$, so

$$x= \frac{\pi}{4} + 2\pi n \space, n \in \Bbb Z \quad\text{and}\quad x= \frac{3\pi}{4} + 2\pi n \space, n \in \Bbb Z $$

are another 2 solutions.

I checked in Wolfram Alpha to see if my work is correct, and I found that there are $10$ solutions to this equation. How can I get the other $6$ solutions I'm missing? (My 4 solutions are correct).


Solution to $\cos \alpha = 0$ is $$\alpha = {\pi \over 2} +\pi n$$

so in your case $$x ={\pi \over 8} +{\pi \over 4}n$$



From $\cos 4x=0$ one gets $4x=\pm\frac\pi 2+2n\pi$, that is $x=\pm\frac{\pi}8+\frac{n\pi}2$. That's eight solutions modulo $2\pi$.

From $\sin x=\frac1{\sqrt2}$ one gets $x=\frac\pi4+2n\pi$ and $x=\frac{3\pi}4+2n\pi$, that two more solutions modulo $2\pi$.

  • $\begingroup$ Why the person that answered before you put one of the solution as $x ={\pi \over 8} +{\pi \over 4}n$ instead of $x ={\pi \over 8} +{\pi \over 2}n$ ? $\endgroup$
    – SilentMath
    Jul 13 '19 at 6:08
  • $\begingroup$ @RodrigoPizarro : Because $-\frac\pi+(2n)\pi=\frac\pi2+(2n-1)\pi$, so that the set of $\pm\frac\pi2+(2n)\pi$ is the same as the set of $\frac\pi2+k\pi$. $\endgroup$ Jul 13 '19 at 7:23

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